**Q.1. **$\lim_{x \to \infty}\frac{x-sinx}{x+cosx}$ equals

**(A) **1
**(B) **-1
**(C) **$\infty$
**(D) **$-\infty$

**Sol : ** $$\lim_{x \to \infty}\frac{x-sinx}{x+cosx} = \lim_{x \to \infty}\frac{1-\frac{sinx}{x}}{1+\frac{cosx}{x}} = 1$$.

Last equality comes from the fact that range of $sinx$ and $cosx$ is [-1,1]. So option **(A)** is correct.

**Q.2. **If P, Q, R are subsets of the universal set U, then $(P\cap Q\cap R) \cup (P^c \cap Q \cap R) \cup Q^c \cup R^c$ is

**(A) **$Q^c \cup R^c$
**(B) **$P \cup Q^c \cup R^c$
**(C) **$P^c \cup Q^c \cup R^c$
**(D) **U

**Sol : **We are given :
$$(P\cap Q\cap R) \cup (P^c \cap Q \cap R) \cup Q^c \cup R^c$$
$$ = (P \cup P^c) \cap (Q \cap R) \cup Q^c \cup R^c \text{ (By distributive law)} $$
$$ = U \cap (Q \cap R) \cup Q^c \cup R^c $$
$$ = (Q \cap R) \cup Q^c \cup R^c $$
$$ = (Q \cap R) \cup (Q \cap R)^c \text{ (By demorgan's law)}$$
$$ = U $$
So option **(D)** is correct.

**Q.3. **The following system of equations
$$x_1 + x_2 + 2x_3 = 1$$
$$x_1 + 2x_2 + 3x_3 = 2$$
$$x_1 + 4x_2 + αx_3 = 4$$
has a unique solution. The only possible value(s) for α is/are

**(A) **0
**(B) **either 0 or 1
**(C) **one of 0, 1, or -1
**(D) **any real number

**Sol : **We write the augmented matrix as :$$\left( \begin{array}{ccc}
1 & 1 & 2 \\
1 & 2 & 3 \\
1 & 4 & α \end{array} \middle\vert \begin{array}{c} 1\\2\\4\end{array}\right)$$
Now we do row operations to bring left matrix to row echelon form.

$$\left( \begin{array}{ccc}
1 & 1 & 2 \\
1 & 2 & 3 \\
1 & 4 & α \end{array} \middle\vert \begin{array}{c} 1\\2\\4\end{array}\right) \xrightarrow{R2\leftarrow R2 - R1, R3\leftarrow R3 - R1}
\left( \begin{array}{ccc}
1 & 1 & 2 \\
0 & 1 & 1 \\
0 & 3 & α-2 \end{array} \middle\vert \begin{array}{c} 1\\1\\3\end{array}\right) \xrightarrow{R3\leftarrow R3 - 3R2}
\left( \begin{array}{ccc}
1 & 1 & 2 \\
0 & 1 & 1 \\
0 & 0 & α-5 \end{array} \middle\vert \begin{array}{c} 1\\1\\0\end{array}\right)$$
So to have unique solution, α can be any real number except 5. So **None** of the options is correct.

**Q.21. **The minimum number of equal length subintervals needed to approximate $\int_1^2 xe^x\,dx$
to an accuracy of at least $\frac{1}{3}*10^{-6}$ using the trapezoidal rule is

**(A) **1000e
**(B) **1000
**(C) **100e
**(D) **100e

**Sol : ** Error in trapezoidal rule is given by :
$$E_n = -\frac{(b-a)^3}{12N^2} f''(c)$$
for some value $c$ between $a$ and $b$.
Now we want maximum arror to be $\frac{1}{3}*10^{-6}$ i.e.
$$\left\vert E_n\right\vert < \frac{1}{3}*10^{-6}$$
We are given $a$ as 1 and $b$ as 2, and we need to find $N$.
By rearranging terms, we get
$$N^2 > \frac{(b-a)^3}{12*\frac{1}{3}*10^{-6}}\left\vert f''(c) \right\vert = \frac{(2-1)^3}{4*10^{-6}}\left\vert f''(c) \right\vert = \frac{10^6}{4}\left\vert f''(c) \right\vert$$
So
$$N > \frac{10^3}{2}\sqrt{\left\vert f''(c) \right\vert}$$
Now $f''(x) = xe^x + 2e^x$, and between 1 and 2, it can take maximum value at 2 (we are looking at max value because we want R.H.S.
of above inequality to be as large as poosible so that we can find least bound on N). So $f''(2) = 4e^2$. So
$$N > \frac{10^3}{2}\sqrt{4e^2} = \frac{10^3}{2}*2e = 1000e$$
So option **(A)** is correct.

**Q.22. **The Newton-Raphson iteration $x_{n+1} = \frac{1}{2}\left(x_n+\frac{R}{x_n}\right)$ can be used to compute the

**(A) **square of R
**(B) **reciprocal of R
**(C) **square root of R
**(D) **logarithm of R

**Sol : ** According to Newton-Raphson method,
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
So we try to bring given equation in above form. Given equation is :
$$x_{n+1} = \frac{x_n}{2}+\frac{R}{2x_n} = x_n - \frac{x_n}{2}+\frac{R}{2x_n} = x_n - \frac{x_n^2-R}{2x_n}$$
So clearly $f(x) = x^2 - R$, so root of f(x) means $x^2 - R = 0$ i.e. we are trying to find square root of R. So option **(C)**
is correct.

See also Q.2.15 (2002), Q.42 (2003), Q.28 (2007) and Q.2 (2010).

**Q.23. **Which of the following statements is true for every planar graph on n vertices?

**(A) **The graph is connected
**(B) **The graph is Eulerian

**(C) **The graph has a vertex-cover of size at most 3n/4
**(D) **The graph has an independent set of size at least n/3

**Sol : **I don't know answer for this. If anyone does, please mail me, I will include that here.

**Q.24. **Let P = $\sum_{\substack{1≤i≤2k \\ i\;odd}} i$ and Q = $\sum_{\substack{1≤i≤2k \\ i\;even}} i$, where k
is a positive integer. Then

**(A) **P = Q - k

**(B) **P = Q + k

**(C) **P = Q

**(D) **P = Q + 2k

**Sol : ** Basically P is sum of first k odd natural numbers, and Q is sum of first k even natural numbers, so clearly each
number in P is 1 less than its corresponding number in Q, and there are k such numbers. So P is k less than Q i.e. P = Q - k.

So option **(A)** is correct.

**Q.25. **A point on a curve is said to be an extremum if it is a local minimum or a local
maximum. The number of distinct extrema for the curve $3x^4-16x^3+24x^2+37$ is

**(A) **0
**(B) **1
**(C) **2
**(D) **3

**Sol : **Extremum of a function can be found by finding zeros of its derivative.

$$f'(x) = 12x^3-48x^2+48x = 0 \rightarrow x(x^2-4x+4) = 0 \rightarrow x(x-2)^2 = 0$$
So extrema lie at $x=0$ and $x=2$. So there are two distinct points of extrema, and so option **(C)** is correct.

**Q.27. **Aishwarya studies either computer science or mathematics everyday. If she
studies computer science on a day, then the probability that she studies
mathematics the next day is 0.6. If she studies mathematics on a day, then the
probability that she studies computer science the next day is 0.4. Given that
Aishwarya studies computer science on Monday, what is the probability that she
studies computer science on Wednesday?

**(A) **0.24
**(B) **0.36
**(C) **0.4
**(D) **0.6

**Sol : **P(CS on wed) = P(CS on tue)*P(CS on wed) + P(Maths on tue)*P(CS on wed) = $(1-0.6)*(1-0.6) + (0.6)*(0.4) = 0.4$

So option **(C)** is correct.

**Q.28. **How many of the following matrices have an eigenvalue 1?

$\left[\begin{array}{cc}1 & 0 \\0 & 0 \end{array} \right]\left[\begin{array}{cc}0 & 1 \\0 & 0 \end{array} \right]
\left[\begin{array}{cc}1 & -1 \\1 & 1 \end{array} \right]$ and $\left[\begin{array}{cc}-1 & 0 \\1 & -1 \end{array} \right]$

**(A) **one
**(B) **two
**(C) **three
**(D) **four

**Sol : **The characteristic equations for given matrices are :
$$-λ(1-λ) = 0$$
$$λ^2 = 0$$
$$(1-λ)(1-λ) + 1= 0$$
$$(-1-λ(-1-λ) = 0$$
λ=1 only satisfies 1st equation. So only first matrix has eigenvalue 1. So option **(A)** is correct.

**Q.29. **Let X be a random variable following normal distribution with mean +1 and
variance 4. Let Y be another normal variable with mean -1 and variance
unknown. If P (X ≤ −1) = P (Y ≥ 2) , the standard deviation of Y is

**(A) **3
**(B) **2
**(C) **$\sqrt{2}$
**(D) **1

**Sol : **We convert random variables X and Y to standard normal variable Z by $Z = \frac{X - μ_x}{σ_x}$ and
$Z = \frac{Y - μ_y}{σ_y}$. So we get

P(Z ≤ $\frac{-1 - 1}{2}) = $P(Z ≥ $\frac{2 + 1}{σ_y})$ i.e.

P(Z ≤ -1) = P(Z ≥ $\frac{3}{σ_y})$

Also we know that, in a standard normal distribution, P(Z ≤ -1) = P(Z ≥ 1).

So $3/σ_y = 1$, so $σ_y = 3$. So option **(A)** is correct.

**Q.31. **P and Q are two propositions. Which of the following logical expressions are
equivalent?

I. P ∨ ~Q

II. ~(~P ∧ Q)

III. (P ∧ Q) ∨ (P ∧ ~Q) ∨ (~P ∧ ~Q)

IV. (P ∧ Q) ∨ (P ∧ ~Q) ∨ (~P ∧ Q)

**(A) **Only I and II
**(B) **Only I, II and III
**(C) **Only I, II and IV
**(D) **All of I, II III and IV

**Sol : **We can't simplify (I) further.

In II, ~(~P ∧ Q) ⇒ P ∨ ~Q, which is I itself.

In III, (P ∧ Q) ∨ (P ∧ ~Q) ∨ (~P ∧ ~Q) ⇒ P ∧ (Q ∨ ~Q) ∨ (~P ∧ ~Q) (By distributive law)
⇒ P ∧ (True) ∨ (~P ∧ ~Q) ⇒ P ∨ (~P ∧ ~Q) ⇒ (P ∨ ~P) ∧ (P ∨ ~Q) ⇒ (P ∨ ~Q).

So III is also equivalent to I.

IV is same as III except last expression is different, and so it is not equivalent to III.

So I, II, and III are equivalent. So option **(B)** is correct.